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I am using the command in R to conduct a test of equality of two proportions:

prop.test(x=c(226,181), n=c(365,335), correct=F)

The p-value varies slightly if I turn on the correct=T. Can anyone tell me that whether I should use correction in this case?

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  • $\begingroup$ correct=TRUE is for considering Yates' continuity correction. $\endgroup$
    – Alexandre Cartaxo
    Dec 7 '15 at 3:14
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    $\begingroup$ Yes, I know. But should I consider continuity correction in this particular case? I am thinking that the n is large so the answer is no... $\endgroup$
    – Sheldon
    Dec 7 '15 at 3:16
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So, Yates showed that the use of Pearson’s chi-squared has the implication of p–values which underestimate the true p–values based on the binomial distribution, but that you already know. Actually, statisticians tend to disagree about whether to use it: some statisticians argue that expected frequency lower that five should imply the use of that correction, while others use ten as that value, while others make the case that Yates' Correction should be used in every chi-squared tests with contingency tables 2 X 2.

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  • $\begingroup$ @Sheldon: I have just run your code and edited the question. The differences in the p-values aren't that small. I should be more conservative in your case and use the Yates' correction stating, nevertheless, the reason of your option. $\endgroup$
    – Alexandre Cartaxo
    Dec 7 '15 at 3:38

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